Energy of a Dipole in a Constant E Field

AI Thread Summary
The discussion focuses on calculating the mechanical and kinetic energy of an electric dipole in a constant electric field, specifically when it oscillates between plus and minus 65 degrees. The potential energy (U) is defined by the equation U = -pEcosθ, with key values provided for specific angles. At 180 degrees, the potential energy is given as 2 μJ, and the user seeks to understand how to apply the equation at 65 degrees. The conversion of potential energy to kinetic energy occurs when the dipole aligns with the electric field at θ = 0. The conversation concludes with a user expressing gratitude for the clarification received on the topic.
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Homework Statement


The graph shows the potential energy of an electric dipole which is in a constant electric field; only the electric force is acting on the dipole. Consider a dipole that oscillates between plus and minus 65 degrees.

knight_Figure_29_10.jpg


a) What is the dipole's mechanical energy?
b) What is the dipole's kinetic energy when it is aligned with the electric field?

Homework Equations


U = -pEcos\theta


The Attempt at a Solution


I know that when the graph is at the trough it's -pE and at it's peak it's +pE. But I don't know how to factor in the 65 degrees. Please help!
 
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At 180 degrees U = 2 μJ. Substitute in the relevant equation and find pE.
At θ = 0, all the potential energy in converted to K.E. Similarly you can find P.E. at 65 degrees by substituting the value of pE and θ in the equation.
 
rl.bhat said:
At 180 degrees U = 2 μJ. Substitute in the relevant equation and find pE.
At θ = 0, all the potential energy in converted to K.E. Similarly you can find P.E. at 65 degrees by substituting the value of pE and θ in the equation.

Thank you so much! That really cleared it up for me!
 
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